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<p><dfn class="terminology">Example 2</dfn> Find the solution of the following initial value problem</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime}=\frac{y \cos x}{1+2y^2}, \quad y(0)=1.
\end{equation*}
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<p class="continuation"><dfn class="terminology">Solution</dfn></p>
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\begin{equation*}
\begin{aligned}
&amp;\frac{\textrm{d} y}{\textrm{d} x}=\frac{\cos x}{\frac{1+2y^2}{y}}~\rightarrow~\frac{1+2y^2}{y} \textrm{d} y=\cos x \textrm{d} x~\rightarrow~\int \frac{1+2y^2}{y} \textrm{d} y=\int \cos x \textrm{d} x~\rightarrow~\ln |y|+y^2=\sin x+C.
\end{aligned}
\end{equation*}
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<p class="continuation">By the initial condition, one has <span class="process-math">\(C=1\)</span> and the solution to the initial value problem is</p>
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\begin{equation*}
\ln |y|+y^2=\sin x+1.
\end{equation*}
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<span class="incontext"><a href="sec2_3.html#p-30" class="internal">in-context</a></span>
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